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Proof step 10 from proof nu-is-UN (
previous step
)
¬(∀ S. compl(f)(S) ⊆ S → S(x)) ⊦ ∃ S. S ⊆ f(S) ∧ S(x)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 11
:
compl(f)(S) ⊆ S ⊦ S(x), ∃ S. S ⊆ f(S) ∧ S(x)
Rule argument
Additional information